# How do you factor completely x^6-1?

May 13, 2016

$\left(x - 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right) \left({x}^{2} + x + 1\right)$
$= \left(x - 1\right) \left(x + 1\right) \left(x - \frac{1}{2} - i \frac{\sqrt{3}}{2}\right) \left(x - \frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(x + \frac{1}{2} - i \frac{\sqrt{3}}{2}\right) \left(x + \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)$

#### Explanation:

The answer in the given form can be obtained directly by observing that it is a cubic in ${x}^{2}$.

${x}^{6} - 1 = {\left({x}^{2}\right)}^{3} - {1}^{3}$

=(x^2-1)((x^2)^2+x^2+!)

$= \left(x - 1\right) \left(x + 1\right) \left({x}^{4} + {x}^{2} + 1\right)$

$= \left(x - 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right) \left({x}^{2} - x + 1\right)$

$= \left(x - 1\right) \left(x + 1\right) \left(x - \frac{1}{2} - i \frac{\sqrt{3}}{2}\right) \left(x - \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)$

$\left(x + \frac{1}{2} - i \frac{\sqrt{3}}{2}\right) \left(x + \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)$

The expanded form can be directly obtained from

$\left({x}^{6} - 1\right) = \prod \left(x - {e}^{i \frac{2 k \pi}{6}}\right)$, k=0, 1, 2,..,5.,

using the six 6th roots of 1, $\left\{{e}^{i \frac{2 k \pi}{6}}\right\}$, k=0, 1, 2,..,5.,